Question

Is this right;

Answer 1

b is correct. for a it's asking where its undefined. meaning when does the denominator equal zero. so factor x^2+3x-10, and set the value to zero to solve.

Answer 2

oh it is (x-5)(x-2)

Answer 3

(x+5)(x-2) now solve individually

Answer 4

your answer is right, @malibugranprix2000 . One vertical asymptote and one discontinuity, and no horizontal asymptote. i'll assume that you have solution to your answers.

Answer 5

oh btw the bottom one is hidden it is what is the oblique asymptote?
I believe the answer is -5

Answer 6

for that question you're gonna have to do polynomial division or synthetic division

Answer 7

I meant for the first one is -5 right?

Answer 8

the oblique asymptote is the line y = x-3. Divide the numerator by the denominator by long division, the quotient is your "oblique" asymptote

Answer 9

x=-5,2 are the vertical asy. x-2=0 is 2

Answer 10

\[\large \frac{ x^3-8 }{ x^2+3x-10 }=(x-3)+\frac{ 19 }{ x+5 }, x \neq 2\]

Answer 11

i assumed he hasn't learned discontinuity yet. but yea x=5 would be it

Answer 12

x=2 is not a vertical asymptote, since \[\frac{ x^3-8 }{ x^2+3x-10 }=\frac{ (x-2)(x^2+2x+4) }{ (x-2)(x+5) }=\frac{ x^2+2x+4 }{ x+5 };x \neq 2\]

Answer 13

Yea x=2 gets cancled right?

Answer 14

yes

Answer 15

oh ok so its not -5 its 5?

Answer 16

that was a mistake when i posted the reply i forgot to put in the neg. sign on 5

Answer 17

oh ok no problem. thanks for the help @technofable and @sirm3d